Posted by **Jordan** on Saturday, December 12, 2009 at 5:59pm.

A person on a small, 7.00 meter high, hill aims a water-balloon slingshot straight at a friend hanging from a tree branch 4.00 meters above and 6.00 meters away. At the instant the water balloon is released at 7.00 m/s, the friend lets go and falls from the tree, hoping to avoid being hit by the water balloon. Calculate the following:

(a) the time for balloon to reach tree ______________s

(b) the height of balloon at that time ________________m

(c) the height of the falling friend at that time ___________________m

i don't even know where to begin. thanks!

- physics =( -
**MathMate**, Saturday, December 12, 2009 at 9:42pm
Draw a picture.

It is not clear to me where the 7m hill comes in. If I interpreted it correctly, it is of no use except that it permits the water balloon and the person to fall freely below the ground!

See:

http://img689.imageshack.us/img689/5825/1260658746b.png

Let

D=horizontal distance between the two friends = 6m

V=vertical distance between the two friends = 4m

u= initial velocity of the balloon making an angle θ (theta) with the horizontal = 7 m/s

ux=horizontal component of initial velocity = ucos(θ)

uy=vertical component of initial velocity = usin(θ)

H1(t)=height above initial level of balloon at time t

= uy*t-(1/2)gt²

H2(t)=height of friend measured from top of tree at time t

= 0 -(1/2)gt²

1. Time to reach tree

T = D/ux

2. H1(T)=uy*T-(1/2)gT²

3. H2(T)=-(1/2)gT²

Note that H1 is measured from ground, and H2 is measured from top of tree, at 4 m above ground.

My calculations tell me that the balloon hits the poor friend.

- physics =) -
**MathMate**, Saturday, December 12, 2009 at 10:08pm
Forgot to mention:

θ=(theta) is the angle aimed directly at the friend, so

θ=arcTan(4m/6m)

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